So I've been fiddling around with the cauchy product of sequences lately, and am curious about a little identity I've found (which I'm sure is ubiquitous in finite differences, as I can't be the only one to have thought of it). I have been having trouble proving it, but the idea is rather straight forward. I think I might have seen it somewhere before but can't find any literature on the subject through searches and am exhausted for a proof (even though I know it should be simple).

Consider a ring of sequences $a:\mathbb{N} \to \mathbb{C}$ endowed with the usual addition $+$ and the multiplication operator $\times$ where

$$(a \times b)(n) = \sum_{j=0}^n a_j b_{n-j}$$

The first identity, which at hand is very simple to prove, follows from associativity of the product. Firstly

$$(a \times 1)(n) = \sum_{j=0}^n a_j$$

and

$$1 \times 1 \times ...(k\,times)...\times 1 = \dbinom{n}{k-1}$$

giving the usual iterated sum formula

$$ \sum^k a(n) = (a\times \dbinom{n}{k-1}) = \sum_{j=0}^n a_j\dbinom{n-j}{k-1}= \sum_{n_{k-1}=0}^n \sum_{n_{k-2}=0}^{n_{k-1}}...k\,times...\sum_{n_0=0}^{n_1} a(n_0)$$

where $\sum^{k}\sum^{j} = \sum^{k+j}$. The question I am curious about is if this has ever been investigated for complex numbers--i.e; does anyone have any references to the operator

$$\sum^{s} a = \sum_{j=0}^n a_j \dbinom{n-j}{s-1}$$

and if perhaps it satisfies the identity

$$\sum^{s} \sum^{q} = \sum^{s+q}$$

so that it forms a sort of differsum (like a differintegral but with sums). By associativity of the Cauchy product the semigroup property is equivalent to showing

$$\sum_{j=0}^n \dbinom{j}{s-1}\dbinom{n-j}{q-1} = \dbinom{n}{s+q-1}$$

which follows for $s,q \in \mathbb{N}$ (by the above) but I am unsure for $\Re(s),\Re(q) > 0$.

This is something I imagine must be somewhere in the vastness of literature on calculus of variations or finite differences. A proof or a reference to a proof is what I'm really looking for. I'm particular to a reference though, as I'd like to read what else people have written on the subject.